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# A coin is tossed $5$ times. Find the probability of getting at least $4$ heads

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## Finding the probability of getting at least $4$ heads:Number of times coin tossed, n $=5$Number of possibilities $=2$ (H, T)Number of possible outcomes $\mathrm{n}\left(\mathrm{S}\right)={2}^{\mathrm{n}}={2}^{5}=32$P(All heads) $={\left(\frac{1}{2}\right)}^{5}=\frac{1}{32}$P(at least $4$ heads)$=\mathrm{p}\left(\mathrm{x}=4\right)+\mathrm{p}\left(\mathrm{x}=5\right)$ $\begin{array}{rcl}\mathrm{P}\left(\mathrm{at}\mathrm{least}4\mathrm{heads}\right)& =& \mathrm{p}\left(\mathrm{x}=4\right)+\mathrm{p}\left(\mathrm{x}=5\right)\\ & ⇒& ={}^{5}\mathrm{C}_{4}{\left(\frac{1}{2}\right)}^{1}{\left(\frac{1}{2}\right)}^{4}+{}^{5}\mathrm{C}_{5}{\left(\frac{1}{2}\right)}^{0}{\left(\frac{1}{2}\right)}^{5}\\ & ⇒& =\frac{5!}{4!\left(5-4\right)!}{\left(\frac{1}{2}\right)}^{1}{\left(\frac{1}{2}\right)}^{4}+\frac{5!}{5!\left(5-5\right)!}{\left(\frac{1}{2}\right)}^{0}{\left(\frac{1}{2}\right)}^{5}\\ & ⇒& =\frac{5!}{4!\left(1!\right)}{\left(\frac{1}{2}\right)}^{1}{\left(\frac{1}{2}\right)}^{4}+\frac{5!}{5!\left(0\right)!}{\left(\frac{1}{2}\right)}^{0}{\left(\frac{1}{2}\right)}^{5}\\ & ⇒& =\frac{5!}{4!}{\left(\frac{1}{2}\right)}^{1}{\left(\frac{1}{2}\right)}^{4}+\frac{5!}{5!}{\left(\frac{1}{2}\right)}^{0}{\left(\frac{1}{2}\right)}^{5}\\ & ⇒& =\frac{5×4×3×2×1}{4×3×2×1}\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^{4}+\frac{5×4×3×2×1}{5×4×3×2×1}{\left(\frac{1}{2}\right)}^{5}\\ & ⇒& =\left(5×\frac{1}{2}×\frac{1}{16}\right)+\left(\frac{1}{32}\right)\\ & ⇒& =\frac{5}{32}+\frac{1}{32}\\ & ⇒& =\frac{6}{32}\\ & ⇒& =\frac{3}{16}\end{array}$Hence, the probability of getting at least $4$ heads is $\frac{3}{16}$.

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